Question

How long will it take to melt 3500 kg of steel in a 50 kW oven operating at 65% efficiency from room temperature to 1600°C?

Answer 1

Answer:

71195 seconds or 19.8 hours

Explanation:

Because the steel will be melted, beginning from room temperature, energy is needed to heat it from room temperature to 1600°C and another to melt it at 1600°C.

We take room temperature to be 27°C, the specific heat capacity of steel to be 420 J/kg/K and the specific latent heat of fusion of steel to be 440 J/kg.

The heat required to heat the steel from room temperature to 1600°C is given by

$H_S = mc_S\delta\theta$

$m$ is the mass, $c_S$ is the specific heat capacity and $\delta\theta$ is the change in temperature.

$H_S = 3500\times420\times(1600-27) = 2312310000$

The heat required to melt it at 1600°C is

$H_L = ml_S$

$m$ is the mass and $l_S$ is the specific latent heat of fusion of steel.

$H_L = 3500\times440 = 1540000$

Adding both, the heat required is $2312310000+1540000=2313850000\text{ J}$

Because the oven is 65% efficient, its output power = $65\%\times50000 = 32500 \text{ W}$

Now, energy = power * time

Time = Energy/power

$t = \dfrac{2313850000}{32500}=71195 \text{ seconds} = 19.8 \text{ seconds}$